(ii) Variation of molar conductivity with concentration for

weak electrolytes :
• The weak electrolytes dissociate to a very lesser extent as compared
to strong electrolytes .
• The molar conductivity is low
as compared to that of
strong electrolytes .

• The variation of
Λm with √c is very large as
compared to that of strong
electrolytes .
• we cannot obtain molar conductance at infinite dilution (Λ0 (by
extrapolation of the Λm versus √c plots .
• The number of free ions of an electrolyte in solution depends upon :
the degree of dissociation with dilution.
• With the increase in dilution, the degree of dissociation increases and as a
result molar conductance increases.
• The limiting value (maximum) of molar conductance (Λ ∞) corresponds to

degree of dissociation equal to 1 “whole of the electrolyte dissociates”.

• Thus, the degree of dissociation α can be calculated at any concentration as,

α= Λ / Λ∞
Λ is the molar conductance at concentration C and Λ ∞ is the molar
conductance at infinite dilution.
❑ The specific conductance increases with increase in concentration of
solution as the number of ions per unit volume increases.
❑ Both the equivalent conductivity and molar conductance increase with
decrease in concentration (i.e. upon dilution) since the extent of
ionization increases.

Explanation:
Since the concentration decreases, one can expect decrease in equivalent conductivity
due to decrease in available number of ions per unit volume.
However the increase in volume (V) factor (increase the mobility) .
Hence the net effect is increase in equivalent conductivity.
3) Temperature:
• The conductivity of an electrolyte depends upon the temperature.
With increase in temperature, the conductivity of an electrolyte
increases.
 Kohlrausch’s Law of
Independent Migration of Ions
• It has been observed that the conductivity of solution
increases with dilution until it reaches its limiting value at
infinite dilution is represented as ∧∞m .
• Kohlrausch made a systematic study of ∧∞m for different
electrolytes and concluded that each ion contributes a
characteristic value of its own to molar conductivity at
infinite dilution irrespective of the nature of the other ion
present.
Electrolytes in set I and II have a
common anion so that the
difference can only be due to
the difference in contribution
to Λ by K+ and Na+ ions.
In the same way in sets III and
IV the constant difference may
be attributed to the difference
in contribution to ∧∞m made
by the Cl- and NO3- ions.
• These observations can be explained by Kohlrausch’s
law of independent migration of ions which states
that:
“At infinite dilution each ion contributes its
definite share to the total equivalent conductivity of
the electrolyte, which depends only on the nature of
the contributing ions and not on the ion with which it
is associated as a part of the electrolyte”.
• In other words,
“The molar conductivity at infinite dilution of an electrolyte
is equal to the sum of the ionic conductances of the ions
composing it, provided the solvent and temperature are the
same”.

∧ = ν + ∧∞ a + ν − ∧∞ c
Where, Λa and Λc are the ionic conductances of the anion and
cation respectively at infinite dilution and ν+ and ν− is the number
of cations and anions in which one molecule of the electrolyte.
Example
Equivalent conductance of NaCl, HCl and C2H5COONa at infinite
dilution are 126.45, 426.16 and 91 ohm–1 cm2 respectively.
Calculate the equivalent conductance of C2H5COOH.

Solution:

λ∞ = λ∞ + λ∞ - λ∞
C2H5COOH C H COONa
2 5 HCL NaCl

= 91 + 426.16 – 126.45
= 390.71 ohm–1 cm2
 ARRHEENIUS THEORY OF
ELECTROLYTIC CONDUCTANCE
Postulates of Arrhenius Theory :
1. When dissolved in water, neutral electrolyte molecules are split up
into two types of charged particles.
• These particles were called ions and the process was ionisation.
• The positively charged particles were called cations and those having
negative charge were called anions.
• The ions are already present in the solid electrolyte held together by
electrostatic force. When placed in water, these neutral
molecules dissociate to form separate anions and cations.
A+ B - A+ + B -
2. The ions present in solution constantly reunit to form neutral
molecules. Thus there is a state of equilibrium between the
undissociated molecules and the ions.
AB < > A + + B-
Applying the Law of Mass Action to the ionic equilibrium we have,

[ A + ][ B − ]
K=
[AB]
where K is called the Dissociation constant.
3. The charged ions are free to move through the solution to the
oppositely charged electrode. This is called as migration of ions.
• This movement of the ions constitutes the electric current
through electrolytes. This explains the conductivity of
electrolytes.
4. The electrical conductivity of an electrolyte solution depends on the
number of ions present in solution.
6. There are two types of electrolytes. Strong electrolytes are those
when dissolved in water are completely dissociated (ionised) into ions.
Like NaCl, KCl, AgNO3 etc.
• In the case of weak electrolytes, there is partial dissociation into ions
in water and an equilibrium exists between the dissociated ions and the
undissociated electrolyte.
CH3COOH < -- --- > CH3COO- + H+.

• The degree of dissociation of an electrolyte determines whether it is a

strong electrolyte or a weak electrolyte.
Evidences of Arrhenius theory of electrolytic dissociation
1. The enthalpy of neutralisation of strong acid by strong base is a
constant value and is equal to -57.32 kJ. gm.equiv -1. This aspect is well
explained by Arrhenius theory of electrolytic dissociation.
• The net reaction in the strong acid-base neutralisation is the formation
of water from H+ and OH- ions.
H+ + OH- -- -- - > H2O, ∆Hro = -57.32 kJ.mol -1

2. The colour of certain salts solution is due to the ions present. For
example, copper sulphate is blue due to Cu2+ ions. Nickel salts are
green due to Ni2+ ions.
3. Chemical reactions between electrolytes are almost ionic reactions.
This is because these are essentially the reaction between oppositely
charged ions. For example,
Ag+ + Cl- --- -- > AgCl
4. Electrolytic solutions conduct current due to the presence of ions
which migrate in the presence of electric field.
5. Colligative properties depend on the number of particles present in
the solution.
• For example, 0.1 molal solution of NaCl has elevation of boiling point
about twice that of 0.1 molal solution of non-electrolyte. The
abnormal colligative properties of electrolytic solutions can be
explained with theory of electrolytic dissociation.
Ostwald's dilution law for weak electrolytes
• According to Arrhenius theory, weak electrolytes partially dissociate
into ions in water which are in equilibrium with the undissociated
electrolyte molecules.
• Ostwald's dilution law relates the dissociation constant of the weak
electrolyte with the degree of dissociation and the concentration of
the weak electrolyte.
• Consider the dissociation equilibrium of CH3COOH
CH3COOH <-- -- > CH3COO- + H+
Initially C 0 0
At equilibrium (1-α)C α α
• the degree of dissociation (α) which represents the fraction of
total concentration of CH3COOH that exists in the ionised state.
• Hence (1 - α) is the fraction of the total concentration of CH3COOH,
that exists in the unionised state. If 'C' is the total concentration of
CH 3COOH initially, then at equilibrium α C, α C and (1 - α) C
represent the concentration of H+, CH3COO- and CH3COOH
respectively.

[ H + ][CH 3COO − ]
Ka =
[CH3COOH]

[αC ][αC ]
Ka =
[(1− α ) C ]
α𝟐 𝑪
Ka = ………………………….. Ostwalds dilution law
(1− α )
If α is too small, then
Ka = α 2C
α = √(Ka/C)
also [H+] = [CH3COO-] = α C
[H+] = √(Ka.C)
For weak bases,
Kb= α 2C / (1- α) and α = √(Kb/C) at α = small values.
Kb = dissociation constant for weak base.
• This law fails for strong electrolytes. For strong
electrolytes, α tends to 1 and therefore Ka increases
tremendously.
• It is applied for weak diluted electrolytes.
 Example
A solution of 0,1 N NaCl has specific conductivity equal to
0.0092.If ionic conductance of Na+ and Cl– ions are 43.0
and 65.0 ohm–1 respectively. Calculate the degree of
dissociation of NaCl solution.

Solution:

Normality=0.10 N

Equivalent conductance of NaCl:

Λ= k x 1000
N
= 0.0092 × 100 = 92 𝑜ℎ𝑚–1

v 92
 =  + +  – = 43 + 65 = 108  = = = 0.85
Na Cl   108
* Example
Calculate α the degree of dissociation and (Ka) the
dissociation constant of 0.01M CH3COOH solution; given o
m
the conductance of CH3COOH is 1.65 ×10 S cm and
–4 –1

(CH3COOH) = 390.6 S cm2 mol–1

Solution :

c K  1000
m =
Molarity

1.65  10 −4  1000
c
m = = 16.5 S cm2mol−1
0.01

=
( c
m )  =
16.5
= 0.042
( m )
o 390
 2
c
Ka =
1− 
0.01 ( 0.042 )
2

1 − 0.042

Ka= 1.84 ×10–5